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In Euclidean spaces, the following maximum principle for minimal surfaces are well known.

Theorem: If $\Sigma_1$, $\Sigma_2 \subset \mathbb{R}^n$ are complete connected minimal hypersurfaces, $\Sigma_1 \cap \Sigma_2 \ne \emptyset$, and $\Sigma_2$ lies on one side of $\Sigma_1$, then $\Sigma_1=\Sigma_2$.

The above theorem stated is Corollary 1.28 in the textbook by Minicozzi and Colding. Though this corollary is stated globally, it's actually a local property. It's a simple consequence of the strong maximum principle of second order elliptic PDEs since every hypersurface can be written as a graph locally.

However, in a Riemannian manifold, there's no canonical way to represent a hypersurface as a graph. My question is:

Question: Does it hold in any Riemannian manifold? Namely, given two open minimal hypersurfaces $\Sigma_1$, $\Sigma_2$ in a Riemannian manifold $\mathcal{M}$. Suppose that $\Sigma_1$, $\Sigma_2$ are contained in a geodesic ball. If $\Sigma_1 \cap \Sigma_2 \ne \emptyset$ and $\Sigma_1$ lies on one side of $\Sigma_2$, does it holds that $\Sigma_1=\Sigma_2$?

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  • $\begingroup$ Yes this is still true. Are you asking for the proof or a reference? $\endgroup$ Commented Dec 29, 2023 at 21:10
  • $\begingroup$ @OtisChodosh Yes, professor. Can you give me a reference? I think maybe I can use Femi coordinates to represent the surfaces as graphs. However, it seems to be a complicated equation which I don't know how to deal with. BTW, I read you lectures "Stable minimal surfaces and positive curvature", that's excellent. $\endgroup$
    – gaoqiang
    Commented Dec 29, 2023 at 21:18
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    $\begingroup$ Nice to hear the notes were useful. I wrote a sketch of the proof below, please feel free to let me know if you want more details anywhere. $\endgroup$ Commented Dec 30, 2023 at 4:10

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I'm not sure of a precise reference, but here is a proof. We work in Fermi coordinates over $\Sigma_1$ and consider $\Sigma_2$ a normal graph over $\Sigma_1$ of some function $f$. We can assume that $f(p) = 0$ and $f\geq 0$. We'll show that $f\equiv 0$ near $p$. By (A.13) here (probably there is a better reference, but we couldn't find one so we worked it out ourselves) we have $$ \textrm{div}_{g_f}\left( \frac{\nabla_{g_f}f}{(1+g_f^{ij}f_if_j)^{1/2}}\right) + \frac{II_{f}^{ij}f_if_j}{(1+g_f^{ij}f_if_j)^{1/2}} - (1+g_f^{ij}f_if_j)^{1/2}H_f = 0 $$ where $g_f = g_{f(y)}+df_y^2$ is the induced metric on the graph of $f$, $II_f$ (resp. $H_f$) is the second fundamental form (resp. mean curvature) of the surface at height $f(y)$.

This is a complicated nonlinear PDE but we claim that $f$ satisfies a simpler PDE that satisfies the maximum principle. To prove this you can argue as in Theorem 11.2 here. Then you get a nice linear elliptic PDE with $Lf=0$. So you can use the standard strong maximum principle.

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